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I know this was already asked before here: Q: The determinant of a matrix $A \in \mathbb{R}^{n \times n}$?

But I still did not manage to solve this with the method he suggested. I tried adding all the columns to the first one (c2+c1, c3+c1.... c n + c1) And then I subtracted the rows from the first one (r2-r1...r3-r1...r n -r1) But after I developed according to the first column I got stuck.

Can anyone guide me or give me an example that shows how to calculate the determinant (step by step)?

Thanks in advance.

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David
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1 Answers1

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Do as written there: add columns $\;2,3,...,n\;$ to column $\;1\;$ and then substract rows $\;2,3,...,n\;$ from first one:

$$\begin{pmatrix}2&1&1&\ldots&1\\ 1&2&1&\ldots&1\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ 1&1&1&\ldots&2\end{pmatrix}\longrightarrow \begin{pmatrix}n+1&1&\ldots&1&1\\ n+1&2&1&\ldots&1\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ n+1&1&1&\ldots&2\end{pmatrix}\longrightarrow$$

$$\begin{pmatrix}n+1&1&\ldots&1&1\\ 0&1&0&\ldots&0\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ 0&0&0&\ldots&1\end{pmatrix}$$

But now you have an upper triangular matrix whose determinant's calculation is trivial.

Timbuc
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  • Now I see where was my mistake. I feel a bit silly asking, but I really didn't quite understand - when you add rows 2,3...n to column 1, I understand why the first number at the first column is n+1. But why do the other numbers at column 1 become n+1? The second number at the 1st column is 1, so you add the numbers of column 2,3,4...n it should be 1+2+(1)^n-1 no? What am I missing? – David Dec 16 '14 at 13:26
  • @EranDavid Take the second row, for example: we add to the first entry all the other entries in that row, getting $$1+2+\overbrace{1+\ldots +1}^{n-2;\text{times}}=1+2+n-2=n+1$$ and similarly with the other first entries in each row. – Timbuc Dec 16 '14 at 13:29
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    That's what I was missing... I understand now. Thank you so much for the explanation! – David Dec 16 '14 at 13:32